Unit 2: Stoichiometry

Introduction to Stoichiometry

Stoichiometry is a fundamental concept in chemistry that allows us to predict the quantities of reactants and products involved in a chemical reaction. Derived from the Greek words "stoicheion" (element) and "metron" (measure), stoichiometry provides the mathematical basis for understanding chemical reactions and is crucial for various applications, from industrial chemical synthesis to everyday laboratory experiments.

1. Dalton's Atomic Theory

In 1808, John Dalton proposed a revolutionary atomic theory that laid the foundation for modern chemistry. His postulates provided a framework for understanding the nature of matter and chemical reactions.

Postulates of Dalton's Atomic Theory:

• Elements are made of atoms: All matter is composed of extremely small, indivisible particles called atoms.
• Atoms are indivisible: Atoms cannot be created, destroyed, or subdivided into smaller particles during chemical reactions.
• Atoms of the same element are identical: Atoms of a given element are identical in mass, size, and all other properties. Atoms of different elements differ in mass, size, and other properties.
• Compounds are combinations of atoms: Compounds are formed when atoms of different elements combine in simple whole-number ratios.
• Chemical reactions involve rearrangement of atoms: Chemical reactions involve the reorganization of atoms, not their creation or destruction. Atoms are merely rearranged, separated, or combined.

Limitations of Dalton's Atomic Theory:

While groundbreaking, Dalton's theory had certain limitations that were later revealed by subsequent discoveries:

• Atoms are divisible (subatomic particles): The discovery of subatomic particles like electrons, protons, and neutrons showed that atoms are not indivisible but are composed of smaller particles.
• Isotopes exist: The discovery of isotopes demonstrated that atoms of the same element can have different masses (due to varying numbers of neutrons), contradicting the postulate that all atoms of the same element are identical in mass. For example, Chlorine has two main isotopes, Cl-35 and Cl-37.
• Not all compounds follow simple ratios: Certain complex compounds, particularly non-stoichiometric compounds (e.g., some metal oxides), do not strictly adhere to simple whole-number ratios of atoms.

2. Laws of Stoichiometry

Several fundamental laws govern the quantitative aspects of chemical reactions, forming the bedrock of stoichiometry.

Law of Conservation of Mass (Antoine Lavoisier, 1789):

"Mass is neither created nor destroyed in any chemical reaction."

This law states that in a closed system, the total mass of the reactants before a chemical reaction must be equal to the total mass of the products after the reaction. Atoms are merely rearranged, not lost or gained. This principle is fundamental to balancing chemical equations.

Example: When 100 g of calcium carbonate (CaCO₃) is heated, it decomposes to produce 56 g of calcium oxide (CaO) and 44 g of carbon dioxide (CO₂). Mass of Reactant (CaCO3) = 100 g Mass of Products (CaO + CO2) = 56 g + 44 g = 100 g The total mass remains conserved.

Law of Definite Proportions (Joseph Proust, 1799):

"A given chemical compound always contains the same elements combined in the same fixed proportion by mass, irrespective of its source or method of preparation."

This law implies that the elemental composition of a pure compound is always constant. For instance, water (H₂O) always consists of hydrogen and oxygen in a 1:8 mass ratio, meaning for every 1 gram of hydrogen, there are 8 grams of oxygen, regardless of whether the water comes from a river or is synthesized in a lab.

Example: Water (H₂O) always contains 11.19% hydrogen and 88.81% oxygen by mass. If you have 100 g of water, it will contain 11.19 g of hydrogen and 88.81 g of oxygen.

Law of Multiple Proportions (John Dalton, 1803):

"When two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element are in ratios of small whole numbers."

This law explains how different compounds can be formed from the same elements. For example, carbon and oxygen can form carbon monoxide (CO) and carbon dioxide (CO₂).

Example: In CO: 12 g of carbon combines with 16 g of oxygen. In CO₂: 12 g of carbon combines with 32 g of oxygen. If we fix the mass of carbon at 12 g, the masses of oxygen that combine with it are 16 g and 32 g. The ratio of these masses of oxygen is 16:32, which simplifies to 1:2, a simple whole-number ratio.

Law of Reciprocal Proportions (Jeremias Richter, 1792):

"When two different elements combine separately with the same fixed mass of a third element, the ratio of the masses in which they do so is either the same as or a simple whole-number multiple of the ratio of the masses in which they combine with each other."

This law is a bit more complex. Consider three elements A, B, and C. If A combines with C, and B combines with C, then the ratio of the masses of A and B that combine with a fixed mass of C will be related to the ratio of masses in which A and B combine directly with each other.

Example: 1. Carbon (C) combines with Hydrogen (H) to form CH₄. Here, 12 g C combines with 4 g H. (Ratio C:H = 3:1) 2. Oxygen (O) combines with Hydrogen (H) to form H₂O. Here, 16 g O combines with 2 g H. (Ratio O:H = 8:1) Let's fix the mass of H at 1 g. Then C combines with 1 g H in a ratio of 3 g C. O combines with 1 g H in a ratio of 8 g O. The ratio of masses of C and O combining with a fixed mass of H (1 g) is 3:8. Now, consider C and O combining with each other to form CO₂. Here, 12 g C combines with 32 g O. The ratio of C:O is 12:32, which simplifies to 3:8. The ratio (3:8) is the same, thus validating the law.

Gay-Lussac's Law of Gaseous Volumes (Joseph Louis Gay-Lussac, 1808):

"When gases react together, they do so in volumes that bear a simple whole-number ratio to one another, and to the volumes of the gaseous products, provided that all volumes are measured at the same temperature and pressure."

This law applies specifically to reactions involving gases and highlights the simple volumetric relationships. It was a crucial precursor to Avogadro's law.

Example: H2(g) + Cl2(g) → 2HCl(g) 1 volume of hydrogen reacts with 1 volume of chlorine to produce 2 volumes of hydrogen chloride. The ratio of volumes is 1:1:2, which is a simple whole-number ratio.

3. Avogadro's Law and Deductions

Building upon Gay-Lussac's work, Amedeo Avogadro proposed a hypothesis that revolutionized the understanding of gases and paved the way for the mole concept.

Avogadro's Law (Amedeo Avogadro, 1811):

"Equal volumes of all gases, at the same temperature and pressure, contain an equal number of molecules."

This law implies that the volume of a gas is directly proportional to the number of moles (or molecules) of the gas, assuming constant temperature and pressure. This explained Gay-Lussac's law of combining volumes and provided a way to determine molecular formulas.

Deductions from Avogadro's Law:

• Molecular mass and Vapour Density:

  Vapour density (VD) is the ratio of the density of a gas to the density of hydrogen gas at the same temperature and pressure. Since, by Avogadro's Law, equal volumes contain equal numbers of molecules, the ratio of densities is also the ratio of their molecular masses.

  Vapour Density (VD) = (Density of gas) / (Density of H2)

  Since density is mass/volume, and volumes are equal for equal numbers of molecules:

  VD = (Mass of 'n' molecules of gas) / (Mass of 'n' molecules of H2)

  VD = (Molecular mass of gas) / (Molecular mass of H2)

  Since the molecular mass of H₂ is 2 (2 x 1 amu), we get:

  VD = (Molecular mass of gas) / 2

  Therefore, the crucial relationship is:

  Molecular mass = 2 x Vapour density

  This formula allows us to determine the molecular mass of a gaseous substance if its vapour density is known.

• Molecular mass and Volume of Gas:

  Avogadro's law leads to the concept of molar volume. At Standard Temperature and Pressure (STP), which is 0°C (273.15 K) and 1 atmosphere (101.325 kPa) pressure, one mole of any ideal gas occupies a fixed volume.

  1 mole of any gas = 22.4 L at STP

  This means that the molecular mass of a gas in grams (molar mass) will occupy 22.4 liters at STP.

• Molecular mass and Number of Particles (Avogadro's Number):

  One mole of any substance (atoms, molecules, ions, etc.) contains a specific number of particles, known as Avogadro's number (N_A).

  1 mole = 6.022 x 1023 particles

  This constant provides a link between the macroscopic world (grams, liters) and the microscopic world (atoms, molecules).

4. Mole Concept

The mole is a central concept in chemistry, providing a convenient way to count very large numbers of atoms, molecules, or other elementary entities. It bridges the gap between the mass of a substance and the number of particles it contains.

Mole:

"A mole is defined as the amount of substance that contains as many elementary entities (atoms, molecules, ions, electrons, or other specified particles) as there are atoms in exactly 12 grams of carbon-12 isotope."

This number of entities is Avogadro's number (N_A), which is approximately 6.022 x 1023.

Relations based on the Mole Concept:

These formulas are essential for performing stoichiometric calculations:

• Number of moles from mass:

  Number of moles (n) = Given mass (m) / Molar mass (M)

  Where:

  • n = number of moles (mol)
  • m = given mass of the substance (g)
  • M = molar mass of the substance (g/mol)

• Number of moles from volume (for gases at STP):

  Number of moles (n) = Given volume (V) / Molar volume (Vm)

  Where:

  • n = number of moles (mol)
  • V = given volume of the gas (L)
  • Vm = molar volume of gas at STP (22.4 L/mol)

• Number of moles from number of particles:

  Number of moles (n) = Number of particles (N) / Avogadro's number (NA)

  Where:

  • n = number of moles (mol)
  • N = given number of particles (atoms, molecules, ions)
  • NA = Avogadro's number (6.022 x 10²³ particles/mol)

• Mass from number of moles:

  Mass (m) = Number of moles (n) x Molar mass (M)

• Volume from number of moles (for gases at STP):

  Volume (V) = Number of moles (n) x 22.4 L/mol (at STP)

Numerical Problem: Mole Calculations

Problem: Calculate the number of moles, number of molecules, and volume at STP for 49 g of H₂SO₄.

Solution:

1. Calculate Molar Mass of H₂SO₄:
   • H: 2 x 1.008 g/mol = 2.016 g/mol
   • S: 1 x 32.06 g/mol = 32.06 g/mol
   • O: 4 x 16.00 g/mol = 64.00 g/mol
   • Molar mass (M) = 2.016 + 32.06 + 64.00 = 98.076 g/mol
2. Calculate Number of Moles (n):

   n = Given mass / Molar mass

   n = 49 g / 98.076 g/mol ≈ 0.50 mol

3. Calculate Number of Molecules (N):

   N = Number of moles x Avogadro's number

   N = 0.50 mol x 6.022 x 1023 molecules/mol

   N = 3.011 x 1023 molecules

4. Calculate Volume at STP (V) (assuming H₂SO₄ is a gas, though it's typically liquid/solid at STP for this context):

   V = Number of moles x Molar volume at STP

   V = 0.50 mol x 22.4 L/mol

   V = 11.2 L

5. Limiting Reactant and Excess Reactant

In most chemical reactions, reactants are not present in exact stoichiometric ratios. One reactant will be completely consumed before the others, thereby limiting the amount of product that can be formed.

• Limiting Reactant:

  The reactant that is completely consumed first in a chemical reaction. It determines the maximum amount of product that can be formed and thus limits the extent of the reaction.

• Excess Reactant:

  The reactant that is left over after the limiting reactant has been completely used up.

Method to Identify the Limiting Reactant:

1. Write and balance the chemical equation.
2. Convert the given masses of all reactants to moles.
3. For each reactant, calculate the amount of product that could be formed if that reactant were completely consumed. This can be done by using mole ratios from the balanced equation.
4. The reactant that produces the smallest amount of product is the limiting reactant. The smallest amount of product calculated is the theoretical yield.

Numerical Problem: Limiting Reactant

Problem: If 10.0 g of hydrogen gas (H₂) reacts with 70.0 g of oxygen gas (O₂) to produce water (H₂O), identify the limiting reactant and calculate the mass of water formed.

Solution:

1. Balanced Chemical Equation:

   2H2(g) + O2(g) → 2H2O(l)

2. Convert masses to moles:
   • Molar mass of H₂ = 2.016 g/mol
   • Molar mass of O₂ = 32.00 g/mol
   • Moles of H₂ = 10.0 g / 2.016 g/mol = 4.96 mol H₂
   • Moles of O₂ = 70.0 g / 32.00 g/mol = 2.19 mol O₂
3. Calculate product (H₂O) formed from each reactant:
   • From H₂:

     According to the balanced equation, 2 moles of H₂ produce 2 moles of H₂O.

     Moles of H2O = 4.96 mol H2 x (2 mol H2O / 2 mol H2) = 4.96 mol H2O

   • From O₂:

     According to the balanced equation, 1 mole of O₂ produces 2 moles of H₂O.

     Moles of H2O = 2.19 mol O2 x (2 mol H2O / 1 mol O2) = 4.38 mol H2O

4. Identify Limiting Reactant and Theoretical Yield:

   Since O₂ produces less H₂O (4.38 mol) compared to H₂ (4.96 mol), Oxygen (O₂) is the limiting reactant.

   The theoretical yield of H₂O in moles is 4.38 mol.

5. Calculate Mass of Water Formed:

   Molar mass of H₂O = 18.016 g/mol

   Mass of H2O = 4.38 mol x 18.016 g/mol = 78.91 g

   Therefore, 78.91 g of water can be formed.

6. Theoretical Yield, Experimental Yield and % Yield

In practical chemistry, the actual amount of product obtained from a reaction often differs from the calculated maximum amount.

• Theoretical Yield:

  The maximum amount of product that can be formed from a given amount of reactants, as calculated from the stoichiometry of the balanced chemical equation. It assumes perfect reaction conditions and 100% conversion of the limiting reactant.

• Experimental Yield (or Actual Yield):

  The amount of product actually obtained from a chemical reaction in the laboratory. This is typically less than the theoretical yield due to various factors like incomplete reactions, side reactions, loss during purification, etc.

• Percentage Yield (% Yield):

  A measure of the efficiency of a chemical reaction, expressed as the ratio of the experimental yield to the theoretical yield, multiplied by 100.

  % Yield = (Experimental yield / Theoretical yield) x 100

Numerical Problem: Percentage Yield

Problem: In an experiment, 15.0 g of ethanol (C₂H₅OH) is reacted with excess oxygen to produce 19.5 g of carbon dioxide (CO₂). Calculate the theoretical yield of CO₂ and the percentage yield for the reaction.

Balanced Equation: C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

Solution:

1. Calculate Molar Masses:
   • C₂H₅OH: (2 x 12.01) + (6 x 1.008) + (1 x 16.00) = 46.068 g/mol
   • CO₂: (1 x 12.01) + (2 x 16.00) = 44.01 g/mol
2. Calculate Moles of Reactant (C₂H₅OH):

   Moles of C2H5OH = 15.0 g / 46.068 g/mol = 0.3256 mol

3. Calculate Theoretical Yield of CO₂ (in moles):

   From the balanced equation, 1 mole of C₂H₅OH produces 2 moles of CO₂.

   Moles of CO2 = 0.3256 mol C2H5OH x (2 mol CO2 / 1 mol C2H5OH) = 0.6512 mol CO2

4. Calculate Theoretical Yield of CO₂ (in grams):

   Theoretical yield of CO2 = 0.6512 mol x 44.01 g/mol = 28.66 g

5. Calculate Percentage Yield:

   Given Experimental yield = 19.5 g

   % Yield = (Experimental yield / Theoretical yield) x 100

   % Yield = (19.5 g / 28.66 g) x 100 = 67.97%

7. Empirical and Molecular Formula from % Composition

The chemical formula of a compound provides information about the elements present and their relative proportions. We can determine these formulas from experimental data, such as percentage composition by mass.

• Empirical Formula:

  The simplest whole-number ratio of atoms present in a compound. It represents the smallest integer ratio of elements.

• Molecular Formula:

  The actual number of atoms of each element in a molecule of the compound. It is a whole-number multiple of the empirical formula.

Steps to Determine Empirical and Molecular Formula:

1. Assume 100g sample and convert percent to grams:

   If the percentage composition is given, assume a 100 g sample of the compound. This allows you to directly convert the percentages into grams for each element.

   e.g., 75% Carbon becomes 75 g Carbon in a 100 g sample.

2. Convert grams to moles:

   Divide the mass of each element (from Step 1) by its atomic mass to find the number of moles of each element.

   Moles = Mass (g) / Atomic mass (g/mol)

3. Find the simplest whole-number ratio:

   Divide the number of moles of each element (from Step 2) by the smallest number of moles calculated. If the ratios are not whole numbers, multiply all ratios by a small integer to obtain whole numbers.

4. Write the empirical formula:

   Use the whole-number ratios from Step 3 as subscripts for each element in the formula.

5. Determine the molecular formula (if molecular mass is known):

   First, calculate the empirical formula mass (sum of atomic masses in the empirical formula).

   Then, determine the integer 'n' by dividing the given molecular mass by the empirical formula mass:

   n = Molecular mass / Empirical formula mass

   Finally, multiply the subscripts in the empirical formula by 'n' to get the molecular formula:

   Molecular formula = n x Empirical formula

Numerical Problem: Empirical and Molecular Formula

Problem: A compound has the following percentage composition: Carbon (C) = 40.0%, Hydrogen (H) = 6.7%, Oxygen (O) = 53.3%. The molecular mass of the compound is 180 g/mol. Determine its empirical and molecular formulas.

Solution:

1. Assume 100 g sample and convert percent to grams:
   • Carbon (C): 40.0 g
   • Hydrogen (H): 6.7 g
   • Oxygen (O): 53.3 g
2. Convert grams to moles (Atomic masses: C=12.01, H=1.008, O=16.00):
   • Moles of C = 40.0 g / 12.01 g/mol = 3.33 mol
   • Moles of H = 6.7 g / 1.008 g/mol = 6.65 mol
   • Moles of O = 53.3 g / 16.00 g/mol = 3.33 mol
3. Find the simplest whole-number ratio:

   Divide by the smallest number of moles (3.33):

   • C: 3.33 / 3.33 = 1
   • H: 6.65 / 3.33 ≈ 2
   • O: 3.33 / 3.33 = 1

   The simplest whole-number ratio is C:H:O = 1:2:1.

4. Write the empirical formula:

   The empirical formula is CH2O.

5. Determine the molecular formula:
   • Calculate Empirical formula mass:

     Empirical formula mass = (1 x 12.01) + (2 x 1.008) + (1 x 16.00) = 12.01 + 2.016 + 16.00 = 30.026 g/mol

   • Calculate 'n':

     Given Molecular mass = 180 g/mol

     n = Molecular mass / Empirical formula mass

     n = 180 g/mol / 30.026 g/mol ≈ 5.99 ≈ 6

   • Write the Molecular formula:

     Molecular formula = n x Empirical formula

     Molecular formula = 6 x (CH2O) = C6H12O6

   The molecular formula of the compound is C6H12O6 (which is glucose).